No. 22 - Turning Number in an Array

No. 22 - Turning Number in an Array

 

Problem: Turning number is the maximum number in an array whichincreases and then decreases. This kind of array is also named unimodal array.Please write a function which gets the index of the turning number in such anarray.

 

For example, theturning number in array {1, 2, 3, 4, 5, 10, 9, 8, 7, 6} is 10, so its index 5is the expected output.

 

Analysis: As we know, the binary search algorithmis suitable to search a number in a sorted array. Since the input array forthis problem is partially sorted, we may also have a try with binary search.

 

Let us try toget the middle number in an array. The middle number of array {1, 2, 3, 4, 5,10, 9, 8, 7, 6} is 5 (the fourth number). It is greater than its previousnumber 4, and less than its next number 10, so it is in the increasingsub-array. Therefore, numbers before 5 can be discarded in the next round ofsearch.

 

The remainingnumbers for the next round of search are {5, 10, 9, 8, 7, 6}, and the number 9is in the middle of them. Since 9 is less than is previous number 10 andgreater than its next number 8, it is in the decreasing sub-array. Therefore,numbers after 9 can be discarded in the next round of search.

 

The remainingnumbers for the next round of search are {5, 10, 9}, and the number 10 is inthe middle. It is noticeable that number 10 is greater than its previous number5 and greater than its next number 9, so it is the maximum number. That is tosay, the number 10 is the turning number in the input array.

 

We can see theprocess above is actually a classic binary search. Therefore, we can implementthe required function based on binary search algorithm, as listed below:

 

int TurningNumberIndex(int* numbers, int length)

{

    if(numbers == NULL || length <= 2)

        return -1;

 

    int left = 0;

    int right = length - 1;

    while(right > left + 1)

   {

        int middle = (left + right) / 2;

        if(middle == 0 || middle == length - 1)

            return -1;

 

        if(numbers[middle] > numbers[middle - 1] &&

           numbers[middle] > numbers[middle + 1])

            return middle;

        else if(numbers[middle] > numbers[middle - 1] &&

           numbers[middle] < numbers[middle + 1])

           left = middle;

        else

           right = middle;

   }

 

    return -1;

}

 

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